# requesting for explanation of surface integration

#### zollen

Evaluate âˆ«âˆ« < x, y, -2 > * N dS, where D is given by z = 1 - x^2 - y^2, x^2 + y^2 <= 1, oriented up.

Cross Product:
$$\displaystyle \int_{0}^{2\pi} \int_{0}^{1} ( r cos \theta , r sin \theta , -2 ) * ( 2r^2 cos \theta , 2r^2 sin \theta , r ) ~ dr d \theta$$

$$\displaystyle \int_{0}^{2\pi} \int_{0}^{1} ( 2r^3 - 2r ) ~dr d \theta = -\pi$$

I managed to get the correct answer. but I didn't understand why the Jacobian of 'r' was missing from the integration since I used polar coordinate.

#### romsek

Math Team
$n dS = (2x,2y,1) \to (2 r \cos(\theta), 2r \sin(\theta), 1)$

for some reason you added a factor of $r$ in which accounts for the missing $r$ in your integral.

#### zollen

$n dS = (2x,2y,1) \to (2 r \cos(\theta), 2r \sin(\theta), 1)$

for some reason you added a factor of $r$ in which accounts for the missing $r$ in your integral.
$$\displaystyle G(r, \theta) = < r cos \theta , r sin \theta , 1 - r^2 >$$
$$\displaystyle dG/dr = < cos \theta , sin \theta , -2r >$$
$$\displaystyle dG/d \theta = < - r sin \theta , r cos \theta , 0 >$$
Cross Product
$$\displaystyle dG/dr ~X~ dG/d \theta = < 2r^2 cos \theta , 2r^2 sin \theta , r >$$

Dot Product
$$\displaystyle \int_{0}^{2\pi} \int_{0}^{1} < r cos \theta , r sin \theta, -2 > * < 2r^2 cos \theta , 2r^2 sin \theta , r >$$

#### romsek

Math Team
$$\displaystyle G(r, \theta) = < r cos \theta , r sin \theta , 1 - r^2 >$$
$$\displaystyle dG/dr = < cos \theta , sin \theta , -2r >$$
$$\displaystyle dG/d \theta = < - r sin \theta , r cos \theta , 0 >$$
Cross Product
$$\displaystyle dG/dr ~X~ dG/d \theta = < 2r^2 cos \theta , 2r^2 sin \theta , r >$$
I really don't think you can start in polar coordinates like that. Not using the standard definition of the surface integral. There is an equivalent definition but the formula is different.

You have to find the normal vector in cartesian coordinates first, dot it with the field vector, then you can convert everything over to polar coordinates for the integration.

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